while in bash script - linux

Question

I am making a bash script. the objective is:
execute a program wait some seconds reset the program and repeat the process.
I make 2 scripts but i don't know where is the error...
#!/bin/bash
while true;
do
seg=`date +%M`;
if [[ "$seg" -eq "30" ]];
then killall sox;
echo "reset";
fi
done
bash: error sintáctico cerca del elemento inesperado `;'
#!/bin/bash
while true;
do
nice -n-10 sox -q -V0 --multi-threaded -t alsa hw:2,0 -t alsa pcm.default &&
done
bash: error sintáctico cerca del elemento inesperado `done'

Answer 1

Issues with Script #1:
The ; notation is to run multiple commands on the same line, one after another. Bash syntax requires that while and do on separate lines (same with if ... and then, and separated by ; if on the same line. Command statements are not normally terminated with a ; char in bash.
Change your code from:
#!/bin/bash
while true;
do
seg=`date +%M`;
if [[ "$seg" -eq "30" ]];
then killall sox;
echo "reset";
fi
done
To:
#!/bin/bash
while true
do
seg=`date +%M`
if [[ "$seg" -eq "30" ]]; then
killall sox
echo "reset"
fi
done
Issues with Script #2:
& means to run the command as a background process. && is used for conditional command chaining, as in: "If the previous command before && succeeds, then run the next command after the &&"
Change from:
#!/bin/bash
while true;
do
nice -n-10 sox -q -V0 --multi-threaded -t alsa hw:2,0 -t alsa pcm.default &&
done
To:
#!/bin/bash
while true
do
nice -n-10 sox -q -V0 --multi-threaded -t alsa hw:2,0 -t alsa pcm.default &
done